Just a reminder that you are arguing with someone who has declared that
the wrong answer is the right one:
Me: "do you still assert that [...] false is the "correct" answer even
though P(P) halts?"
PO: Yes that is the correct answer even though P(P) halts.
On 6/19/2023 3:08 PM, Ben Bacarisse wrote:
Just a reminder that you are arguing with someone who has declared that the wrong answer is the right one:
Me: "do you still assert that [...] false is the "correct" answer even
though P(P) halts?"
PO: Yes that is the correct answer even though P(P) halts.
Because
*Ben Bacarisse targets my posts to discourage honest dialogue*
*Ben Bacarisse targets my posts to discourage honest dialogue*
*Ben Bacarisse targets my posts to discourage honest dialogue*
All of my posts will be entitled as a Rebuttal to Ben
It is an easily verified fact that P correctly simulated by H cannot
possibly reach its own last instruction and terminate normally thus from
the Professor Sipser agreed criteria the input to H(P,P) does not halt.
MIT Professor Michael Sipser has agreed that the following verbatim
words are correct (he has not agreed to anything else):
(a) If simulating halt decider H correctly simulates its input D until H correctly determines that its simulated D would never stop running
unless aborted then
(b) H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
To address what looks like a contradiction to reviewers not having a
very deep understanding of the halting problem:
(1) A return value of 1 from H(D,D) means the input to H(D,D) has halted
(2) A return value of 0 from H(D,D) has been redefined to mean
(a) D does not halt
(b) D has been defined to do the opposite of whatever Boolean value
that H returns.
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
*Now for the new material*
I am specifically defining the set of “halting problem” finite string pair instances such that TMD2 does the opposite of the Boolean value
that each element of TMD1 returns.
The above set is the set where the behavior of the directly executed TMD2(TMD2) is out-of-sync with the return value of TMD1(TMD2,TMD2)
*ONLY* because TMD2 has been defined to contradict whatever Boolean
value that TMD1 returns.
This makes each {TMD1, TMD2} pair isomorphic to the Liar Paradox and the
Liar Paradox: "This sentence is not true" is an unsound statement, thus
the question: Is the Liar Paradox true or false becomes an unsound
question.
Thus the question: "Does this input halt?"
is isomorphic to this question:
Is the Liar Paradox true or false?
For every {TMD1, TMD2} pair defined above.
We can know in advance that every Boolean return value from every
element of the set of TMD1 deciders necessarily out-of-sync with the
behavior of its corresponding TMD2 input because every element of the
{TMD1, TMD2} pairs has been defined to have that property.
When anyone says that we have to wait and see it is obvious that they
are only playing deceptive head games.
On 6/19/2023 3:08 PM, Ben Bacarisse wrote:
Just a reminder that you are arguing with someone who has declared that the wrong answer is the right one:
Me: "do you still assert that [...] false is the "correct" answer even
though P(P) halts?"
PO: Yes that is the correct answer even though P(P) halts.
Because
*Ben Bacarisse targets my posts to discourage honest dialogue*
*Ben Bacarisse targets my posts to discourage honest dialogue*
*Ben Bacarisse targets my posts to discourage honest dialogue*
All of my posts will be entitled as a Rebuttal to Ben
It is an easily verified fact that P correctly simulated by H cannot
possibly reach its own last instruction and terminate normally thus from
the Professor Sipser agreed criteria the input to H(P,P) does not halt.
MIT Professor Michael Sipser has agreed that the following verbatim
words are correct (he has not agreed to anything else):
(a) If simulating halt decider H correctly simulates its input D until H correctly determines that its simulated D would never stop running
unless aborted then
(b) H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
To address what looks like a contradiction to reviewers not having a
very deep understanding of the halting problem:
(1) A return value of 1 from H(D,D) means the input to H(D,D) has halted
(2) A return value of 0 from H(D,D) has been redefined to mean
(a) D does not halt
(b) D has been defined to do the opposite of whatever Boolean value
that H returns.
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
THIS CHANGE UTTERLY REFUTES BEN'S REBUTTAL
*Now for the new material*
I am specifically defining the set of “halting problem” finite string pair instances such that TMD2 does the opposite of the Boolean value
that each element of TMD1 returns.
The above set is the set where the behavior of the directly executed TMD2(TMD2) is out-of-sync with the return value of TMD1(TMD2,TMD2)
*ONLY* because TMD2 has been defined to contradict whatever Boolean
value that TMD1 returns.
This makes each {TMD1, TMD2} pair isomorphic to the Liar Paradox and the
Liar Paradox: "This sentence is not true" is an unsound statement, thus
the question: Is the Liar Paradox true or false becomes an unsound
question.
Thus the question: "Does this input halt?"
is isomorphic to this question:
Is the Liar Paradox true or false?
For every {TMD1, TMD2} pair defined above.
We can know in advance that every Boolean return value from every
element of the set of TMD1 deciders necessarily out-of-sync with the
behavior of its corresponding TMD2 input because every element of the
{TMD1, TMD2} pairs has been defined to have that property.
When anyone says that we have to wait and see it is obvious that they
are only playing deceptive head games.
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